Answer:
No. It's 15,609
Step-by-step explanation:
The fifteen thousand represents the 15,000
The six hundred is 600
The nine is 9.
So it is 15,609
We conclude that the solution to the linear equation is x = 1/24.
<h3>
How to find the number?</h3>
Let's define x as the number.
Now we want to write:
"the sum of 2/3 and for times a number...."
That can be written with the equation:
2/3 + 4*x
And we also know that:
"the sum of 2/3 and for times a number is equal to 5/6"
Then we can complete the linear equation:
2/3 + 4*x = 5/6
Now we need to solve it for x:
2/3 + 4x = 5/6
4x = 5/6 - 2/3
4x = 5/6 - 4/6
4x = 1/6
x = (1/4)*(1/6) = 1/24
x = 1/24
We conclude that the solution to the linear equation is x = 1/24.
If you want to learn more about linear equations:
brainly.com/question/1884491
#SPJ1
Answer:
- 8
Step-by-step explanation:
The common difference d between consecutive terms is
d = 
- 
, that is
d = - 4 - (- 2) = - 4 + 2 = - 2
To obtain the next term subtract - 2 from - 6
next term = - 6 - 2 = - 8
Is three points that divide sorted data set<span> into four equal groups (by count of numbers), each representing a fourth of the distributed sampled population. There are three </span>quartiles..
<span>Hope That Helped =D</span>
Complete question is;
A researcher wished to compare the average amount of time spent in extracurricular activities by high school students in a suburban school district with that in a school district of a large city. The researcher obtained a simple random sample of 60 high school students in a large suburban school district and found the mean time spent in extracurricular activities per week to be six hours with a standard deviation of three hours. The researcher also obtained an independent simple random sample of 40 high school students in a large city school district and found the mean time spent in extracurricular activities per week to be four hours with a standard deviation of two hours. Let x¯1 and x¯2 represent the mean amount of time spent in extracurricular activities per week by the populations of all high school students in the suburban and city school districts, respectively. Assume two-sample t procedures are safe to use?
what is the 95% confidence interval a researcher wishes to compare the average amount of time spent in extracurricular?
Answer:
CI = (0.755, 3.245)
Step-by-step explanation:
For SRS of 60;
Mean: x1¯ = 6
Standard deviation: s1 = 3
For SRS of 40;
Mean: x2¯ = 4
Standard deviation; s2 = 2
Critical value for the confidence interval of 95% is: t = 1.96
Formula for the CI is;
CI = (x¯1 - x¯2) ± t√[(s1²/n1) + ((s2)²/n1)]
Plugging in the relevant values, we have:
CI = (6 - 4) ± 1.96√[(3²/60) + ((4)²/40)]
CI = 2 ± 1.96√[(3²/60) + ((4)²/40)]
CI = 2 ± 1.96√0.55
CI = 2 ± 1.245
CI = [(2 - 1.245), (2 + 1.245)]
CI = (0.755, 3.245)
